Three Dimensional Geometry — Class 12 Maths Solution

. : (a) The given plane is $z = 2$

...(1)
Direction cosines of the normal to the plane are $< 0,0,1 >$

And distance of (1) from the origin $= \cfrac{{|0 - 2|}}{{\sqrt {0 + 0 + 1} }} = \cfrac{2}{1} = 2$

(b) The given plane is $x + y + z = 1$

...(1)
Direction-ratios of the normal to the plane are $< 1,1,1 >$

therefore Direction cosines of the normal to the plane are $< \cfrac{1}{{\sqrt 3 }},\cfrac{1}{{\sqrt 3 }},\cfrac{1}{{\sqrt 3 }} >$

And distance of (1) from the origin $= \cfrac{{|0 + 0 + 0 - 1|}}{{\sqrt {1 + 1 + 1} }} = \cfrac{1}{{\sqrt 3 }}$

(c) The given plane is $2x + 3y - z = 5$

...(1)
Direction-ratios of the normal to the plane are$< 2,3, - 1 >$

therefore Direction cosines of the normal to the plane are :

$< \cfrac{2}{{\sqrt {4 + 9 + 1} }},\cfrac{3}{{\sqrt {4 + 9 + 1} }},\cfrac{{ - 1}}{{\sqrt {4 + 9 + 1} }} > i.e. < \cfrac{2}{{\sqrt {14} }},\cfrac{3}{{\sqrt {14} }},\cfrac{{ - 1}}{{\sqrt {14} }} >$

And distance of (1) from the origin $= \cfrac{{|0 + 0 - 0 - 5|}}{{\sqrt {4 + 9 + 1} }} = \cfrac{5}{{\sqrt {14} }}$

(d) The given plane is $5y + 8 = 0$

...(1)
Direction ratios of the normal to the plane are $< 0,5,0 >$ i.e., $< 0,1,0 >$

therefore Direction cosines of the normal to the plane are $< 0,1,0 >$

And distance of (1) from the origin $= \cfrac{{|0 + 8|}}{{\sqrt {0 + 25 + 0} }} = \cfrac{8}{5}$